Least Common Multiples, modern and ancient

Three 2001 Russian encyclopedia entries, a 2008 blog, and a 2008 Wikipedia entry point out Ahmes' 1650 BCE red auxiliary numbers as a central least common multiple (LCM) theme. The LCM theme reports modern and ancient terms of a unifying theme of ancient Egyptian arithmetic, rather than a secondary fragmented bit of information. Ahmes' LCMs had been under read by 1920s math historians. After 2002 close reviews of 2/n tables and other information, Ahmes has been shown to have selected optimized, but not always optimal, LCMs to write out 2/n table and calculate other optimized Egyptian fraction series. The Russian encyclopedia entries are published on-line by Springer with the first entry being:

http://eom.springer.de/a/a011920.htm

The first entry suggests that a LCM form may have been known to Ahmes by writing out

3/11 = ! 1/6 + 1/11 + 1/66

It is highly likely that Ahmes shorthand included:

3/11 times (6/6) = (18/66) = (11 + 6+ 1)/66 = 1/6 + 1/11+ 1/66

As many readers of Egyptian math history are aware Ahmes' red auxiliary numbers was an idea that has long been associated with LCMs. The precise nature of LCMs has come to light.

The second Russian entry only mentions modern LCMs per:

http://eom.springer.de/F/f041200.htm

and therefore the entry did not hypothetically offer an ancient leap back in time to decode the 2/n table. Had a formal LCM decoding hypothesis of the RMP 2/n table been offered, academic discussions may have been established within an interdisciplinary context. Following the modern logic offered by the Russian encyclopedia entries, formal proofs, or refutations could be discussed.

Four ways to di! scuss formal Egyptian proofs, taken from

http://ahmespapyrus.blogspot.com/2009/01/ahmes-papyrus-new-and-old.html ,


that Russians had not considered are summarized by:

1. Read/translte RMP 38, that solved the problem:

a. 10 hekat equals 3200 ro times 7/22 equals 101 9/11

b. proving 101 9/11 times 22/7 equals 3200 ro

c. with a footnote that 35/11 times 1/10 = 35/110 = 7/22

2. Read/translate RMP 47, that solved the problem

a. 100 hekat equals 6400/64 hekat divided by 70

b. proving (91/64)hekat + 150/70 ro was the answer

3. Read/translate RMP 66 problem

a. 10 hekat times 32o ro equals 3200 ro, divided by 365, the number of days per year

b. 3200/365 equals 8 + 280/365 with

c. a duplation proof for the quotient 8, by writing

365 1
730 2
1460 4
2920 8

and the remainder (3200 - 2920) ! equals 280 by

(243 1/3 + 36 1/2 + 1/6)/365 = 280/365

4. RMP 82 an began with a hekat unity, (64/64), divided by 29 divisors n in the range 1/64 < n < 64, as RMP 83 divided (64/64), divided by 6, 20 , and 40.


A final Russian entry generally used an ancient aliquot fraction or ratio idea (Russian terms for red auxiliary numbers) without identifying a hypothetical use of ancient LCMs per:

http://eom.springer.de/A/a013260.htm

Note the straight forward modern and ancient arithmetic that easily fills in Ahmes use of red auxiliary LCMs per:

http://rmprectotable.blogspot.com/

and,

http://en.wikipedia.org/wiki/Red_auxiliary_numbers

math least common multiple